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Question

An insulated container of two diatomic gases has two chambers separated by an insulting partition. One of the chambers has volume V1 and contains ideal gas at pressure p1 and temperature T1. The other chamber has volume V2 and contains ideal gas at pressure p2 and temperature T2. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be


A

T1T2(p1V1+p2V2)p1V1T1+p2V2T2

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B

p1V1T2+p2V2T1p1V1+p2V2

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C

p1V1T1+p2V2T2p1V1+p2V2

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D

T1T2(p1V1+p2V2)p1V1T2+p2V2T1

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Solution

The correct option is D

T1T2(p1V1+p2V2)p1V1T2+p2V2T1


As no work is done and system is thermally insulated from surrounding, it means sum of internal energy of gas in two partitions is constant i.e., U=U1+U2. Assuming both gases have same degree of freedom, then
U=f(n1+n2)RT2 and U1=f(n1RT1)2,U2=fn2RT22

Solving we get, T=(p1V1+p2V2)T1T2p1V1T2+p2V2T1


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